The molar solubility of lead iodide is 1.45*10^-3M at 20 degrees C and 6.5*10^-3M at 80 degrees C .
What are the values of delta H not and delta S not for dissolution of PbI2 ?
PbI2(s) ---> Pb2+(aq) + 2I- (aq)
You may assume that delta H and delta S are independent of temperature.
To find the values of ΔH° and ΔS° for the dissolution of PbI2, we can use the Van't Hoff equation, which relates the equilibrium constant (K) to the temperature dependence of ΔH° and ΔS°.
The Van't Hoff equation is given as:
ln(K2/K1) = (-ΔH°/R) × (1/T2 - 1/T1)
Where:
K2 and K1 are the equilibrium constants at two different temperatures
ΔH° is the standard enthalpy change
R is the ideal gas constant (8.314 J/(mol K))
T2 and T1 are the temperatures at which K2 and K1 are measured respectively.
In this case, we are given the molar solubility of PbI2 at two temperatures: 20°C (293 K) and 80°C (353 K). Since the molar solubility is directly related to the equilibrium constant, we can say:
K2 = [Pb2+][I-]^2 at 80°C
K1 = [Pb2+][I-]^2 at 20°C
Now let's plug the values into the equation:
ln([Pb2+][I-]^2 at 80°C / [Pb2+][I-]^2 at 20°C) = (-ΔH°/R) × (1/353 K - 1/293 K)
Since the molar solubility values are given, we can use these values as concentration terms. The solubility of Pb2+ and I- ions at equilibrium would be equal to the molar solubility of PbI2, which is given as 1.45 × 10^-3 M at 20°C and 6.5 × 10^-3 M at 80°C.
So we have:
ln((1.45 × 10^-3 M × (6.5 × 10^-3 M)^2) / (1.45 × 10^-3 M × (6.5 × 10^-3 M)^2)) = (-ΔH°/R) × (1/353 K - 1/293 K)
Now we can simplify the equation:
ln(1) = (-ΔH°/R) × (1/353 K - 1/293 K)
Since ln(1) = 0, we have:
0 = (-ΔH°/R) × (1/353 K - 1/293 K)
Now we can solve for ΔH°:
ΔH° = R × (1/353 K - 1/293 K)
Using the ideal gas constant R = 8.314 J/(mol K), we can substitute the values to calculate ΔH°.
Similarly, we can use the same equation to find ΔS°. The new equation will be:
ΔS° = R × ln(K)
Where K is the equilibrium constant at a specific temperature.
Given that we have already calculated ΔH°, we can use the equation:
ΔS° = R × ln(K)
Substituting in the equilibrium constant at 20°C or 80°C, we can calculate ΔS° for dissolution of PbI2.
Please note that this assumes that the temperature dependence of ΔH° and ΔS° is independent, as specified in the question.